NCERT Solutions for Class 9th Maths Chapter 7 Chapter 7 review (questions marked * are the harder set) — End-of-Chapter Exercises

Book page 169–173 Updated on2026-09-08

Q1.

Fill in the blanks. (i) The probability of an impossible event is _______. (ii) The set of all possible outcomes of a random experiment is called the __________. (iii) The probability of an event that is certain to happen is _______. (iv) Tossing a fair coin has a probability of ______ for getting heads.

Answer

Part	Answer	Reason
(i) Probability of an impossible event	0	No outcome is favourable, so P = 0/n(S) = 0.
(ii) Set of all possible outcomes	sample space (S)	By definition; its size is the sample size n(S).
(iii) Probability of a certain event	1	Every outcome is favourable, so P = n(S)/n(S) = 1.
(iv) Heads on a fair coin	1/2 (0.5 or 50%)	S = {H, T}, both equally likely, one favourable.

Why it happens: (i) and (iii) are the two ends of the probability scale, and they explain why no probability can lie outside it. The favourable count n(A) can never be smaller than 0 or larger than n(S), so the fraction n(A)/n(S) is trapped between 0 and 1 — which is the statement 0 ≤ P(E) ≤ 1 in the Chapter Summary.

Tip: P(E) = 0 and P(E) = 1 are opposites in a precise sense: if E is impossible then 'not E' is certain, and P(not E) = 1 − P(E) = 1 − 0 = 1.

Q2.

In a survey of 50 students, 15 students said they liked football. The number of students who like football is 15, and the ________ (frequency/relative frequency) is __________ (fill in the fraction or decimal).

Answer

The blanks are filled by relative frequency and 15/50 = 3/10 = 0.3.

Frequency of 'likes football' = 15 (a count)
Relative frequency = 15/50 = 3/10 = 0.3 or 30% (a fraction of the total)

Why it happens: the sentence deliberately puts the two words side by side. Frequency answers "how many?" and is a whole number that grows as you survey more people. Relative frequency answers "what share?" — it divides by the total, so it always lies between 0 and 1 and can be compared across surveys of different sizes. It is the relative frequency, not the frequency, that serves as an estimate of probability: P(a randomly chosen student likes football) ≈ 0.3.

Tip: a second survey of 200 students might give a frequency of 62 — a much bigger number — yet a relative frequency of 62/200 = 0.31, almost the same. That stability is exactly why probability is defined as a proportion.

Q3.

Which of the following experiments have equally likely outcomes? Explain. (i) A driver attempts to start a car. The car starts or does not start. (ii) Tossing a fair coin once. (iii) Rolling a fair 6-sided die. (iv) Choosing a marble randomly from a bag that contains 3 red marbles and 7 blue marbles. (v) A baby is born. It is a boy or a girl.

Answer

Only (ii) and (iii) have equally likely outcomes. (v) is treated as equally likely in school problems but is not exactly so.

Experiment	Equally likely?	Explanation
(i) Car starts / does not start	No	The result depends on the battery, the fuel and the age of the car — physical facts, not symmetry. A well-serviced car starts far more often than not.
(ii) Tossing a fair coin once	Yes	The coin is symmetrical and unbiased, so there is no reason to favour either face: P(H) = P(T) = 1/2.
(iii) Rolling a fair 6-sided die	Yes	All six faces are identical in shape and weight, so each has probability 1/6.
(iv) A marble from 3 red and 7 blue	No (by colour)	P(red) = 3/10 and P(blue) = 7/10. The ten marbles are equally likely at 1/10 each; the two colours are not.
(v) A baby is a boy or a girl	Very nearly, but not exactly	Birth records worldwide show slightly more boys than girls — roughly 51 boys per 100 births. School problems model it as 1/2 each, which is a good approximation, not a fact of symmetry.

Why it happens: two outcomes being the only two outcomes does not make them equally likely — that is the mistake (i), (iv) and (v) are all testing. Equal likelihood has to come from somewhere: a physical symmetry, as in the coin and the die, or from data. In (iv) the symmetry is real but it belongs to the marbles, not to the colours, which is why relabelling the sample space as {M₁, …, M₁₀} makes the counting formula work again.

Tip: whenever a problem offers you exactly two outcomes, resist the pull of 1/2. Ask what makes them equal. In (i) nothing does.

Q4.

Write the sample space and calculate the probability based on the given information. (i) Two coins are tossed at the same time. What is the probability of getting at least one head? (ii) Ten identical cards numbered 1 to 10 are placed in a box. One card is drawn at random. What is the probability of drawing a card with an even number? (iii) A die is rolled once. What is the probability of getting a number greater than 4? (iv) A bag contains 3 red balls, 2 blue balls, and 1 green ball. One ball is picked at random. What is the probability that it is not red? (v) Three coins are tossed simultaneously. What is the probability of getting exactly two heads?

Answer

(i) Two coins, at least one head.

S = {HH, HT, TH, TT}, n(S) = 4
E = {HH, HT, TH}, n(E) = 3
P(at least one head) = 3/4 = 0.75

Or by the complement: the only way to fail is TT, so P = 1 − 1/4 = 3/4.

(ii) One card from 1 to 10, even number.

S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, n(S) = 10
E = {2, 4, 6, 8, 10}, n(E) = 5
P(even) = 5/10 = 1/2 = 0.5

(iii) A die, number greater than 4.

S = {1, 2, 3, 4, 5, 6}, n(S) = 6
E = {5, 6}, n(E) = 2
P(greater than 4) = 2/6 = 1/3 ≈ 0.333

(iv) 3 red, 2 blue, 1 green — not red. Label the balls so that all six are equally likely:

S = {R₁, R₂, R₃, B₁, B₂, G}, n(S) = 6
E = 'not red' = {B₁, B₂, G}, n(E) = 3
P(not red) = 3/6 = 1/2 = 0.5

Check with the complement: P(red) = 3/6 = 1/2, so P(not red) = 1 − 1/2 = 1/2. ✓

(v) Three coins, exactly two heads. Each coin doubles the list, so n(S) = 2 × 2 × 2 = 8.

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}, n(S) = 8
E = 'exactly two heads' = {HHT, HTH, THH}, n(E) = 3
P(exactly two heads) = 3/8 = 0.375

Why it happens: in every part the working is the same two steps — write a list whose members are equally likely, then count. The care goes into the list. In (iv) the colours are not equally likely, so the balls are labelled first; in (v) 'exactly two heads' has to be split into the three orders in which it can occur, because HHT, HTH and THH are three different outcomes of the experiment even though they look alike in words.

Tip: 'at least one' in (i) and 'not red' in (iv) are both easier through the complement, P(not E) = 1 − P(E). With three coins, 'at least one head' would be 1 − P(TTT) = 1 − 1/8 = 7/8 — one line instead of listing seven outcomes.

Q5.

A bag has 3 candies: strawberry, lemon, and mint. One is picked at random. What is the probability of picking a strawberry candy?

Answer

S = {strawberry, lemon, mint}, n(S) = 3
E = {strawberry}, n(E) = 1
P(strawberry) = 1/3 ≈ 0.333 or about 33.3%

Why it happens: here — unlike the bags of coloured balls elsewhere in this exercise — there is exactly one candy of each kind, so the three outcomes really are equally likely and 1/3 is honest. The phrase "at random" is what guarantees it: no candy is easier to reach than another.

Tip: change the bag to 3 strawberry, 1 lemon and 1 mint and the answer becomes 3/5, not 1/3, even though there are still three flavours. Count candies, not names.

Q6.

A child has 2 shirts (one red and one blue) and 3 types of pants (jeans, khakis, and shorts). List all the possible combinations of outfits consisting of one shirt and one pair of pants. Display your answer in a table format.

Answer

Every shirt can go with every pair of pants, so there are 2 × 3 = 6 outfits.

Shirt \ Pants	Jeans	Khakis	Shorts
Red	Red + Jeans	Red + Khakis	Red + Shorts
Blue	Blue + Jeans	Blue + Khakis	Blue + Shorts

S = {(Red, Jeans), (Red, Khakis), (Red, Shorts),
(Blue, Jeans), (Blue, Khakis), (Blue, Shorts)}
n(S) = 2 × 3 = 6

Why it happens: the two choices are made together, not instead of each other, so the counts multiply. Each row of the table is one shirt paired with all three pants; two rows of three cells give six cells. This is the same multiplication that gave 6 outcomes for the die-and-coin pair in Exercise Set 7.3 and 4 outcomes for two coin tosses in Fig. 7.6.

Tip: if the child dressed at random, each cell would have probability 1/6, so P(wearing shorts) = 2/6 = 1/3 — the shorts column. A table is a sample space laid out in a rectangle.

Q7.

A tyre company records distances before replacement in 1000 cases. Distance (km): Less than 4000 — 20 cases; 4001 to 9000 — 210 cases; 9001 to 14000 — 325 cases; More than 14000 — 445 cases. Find the probability that a randomly chosen tyre lasts: (i) Less than 4000 km. (ii) Between 4000 and 14000 km. (iii) More than 14000 km.

Answer

Total cases: 20 + 210 + 325 + 445 = 1000 ✓ — so each probability is (cases in that class) ÷ 1000.

Distance (km)	Less than 4000	4001 to 9000	9001 to 14000	More than 14000
Number of cases	20	210	325	445
Probability	0.02	0.21	0.325	0.445

(i)

P(less than 4000 km) = 20/1000 = 1/50 = 0.02 or 2%

(ii) 'Between 4000 and 14000' covers the two middle classes together:

Favourable cases = 210 + 325 = 535
P = 535/1000 = 107/200 = 0.535 or 53.5%

(iii)

P(more than 14000 km) = 445/1000 = 89/200 = 0.445 or 44.5%

Check: 0.02 + 0.535 + 0.445 = 1 ✓ — the three answers cover every one of the 1000 cases exactly once.

Why it happens: these are experimental probabilities read from real records, not theoretical ones. There is no symmetry among the four distance classes — a tyre is far likelier to last past 14000 km than to fail before 4000 km — so the numbers must come from the data. What licenses the division by 1000 is that one case out of the thousand is chosen at random, so every recorded case is equally likely to be the one picked.

Tip: the printed classes are 'less than 4000' and '4001 to 9000', so a tyre lasting exactly 4000 km falls in no class. The table treats the distances as whole kilometres and part (ii) asks about the two middle classes, which is how the 535 arises. In your own tables, write classes that leave no gaps.

Q8.

The letters of the word ‘PEACE’ are placed on cards. Leela draws a card without looking. (i) What is the probability that it is a P, E or C? (ii) What is the probability that it is not an E?

Answer

PEACE has 5 letters, so there are 5 cards: P, E, A, C, E. Each card is equally likely, so n(S) = 5. Note that E appears on two cards.

Letter	P	E	A	C
Number of cards	1	2	1	1
Probability	1/5	2/5	1/5	1/5

(i) Favourable cards: P (1 card), E (2 cards) and C (1 card).

n(E) = 1 + 2 + 1 = 4
P(P, E or C) = 4/5 = 0.8 or 80%

(ii) The cards that are not E are P, A and C.

n(not E) = 5 − 2 = 3
P(not E) = 3/5 = 0.6 or 60%
Check: P(E) + P(not E) = 2/5 + 3/5 = 1 ✓

Why it happens: the repeated E is the whole point of choosing this word. There are 4 different letters but 5 cards, and it is the cards that are drawn at random and therefore equally likely. Answering (i) with 3/4 — three chosen letters out of four distinct letters — ignores that the second E is a real card that Leela can pull out.

Tip: part (ii) is quicker by the complement rule: P(not E) = 1 − P(E) = 1 − 2/5 = 3/5. Also notice (i) and (ii) are related — 'P, E or C' and 'not E' overlap but are not opposites, so you cannot get one from the other.

Q9.

*A game of chance consists of spinning an arrow (see Fig. 7.7.) which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, and these are equally likely outcomes. What is the probability that it will point at (i) 8? (ii) An odd number? (iii) A number greater than 2? (iv) A number less than 9? (v) A multiple of 3?

Answer

The wheel of Fig. 7.7 is cut into 8 equal sectors numbered 1 to 8, and the question tells us they are equally likely. So n(S) = 8 with S = {1, 2, 3, 4, 5, 6, 7, 8}, and every answer is (how many sectors qualify) ÷ 8.

Part	Favourable sectors	Count	Probability
(i) 8	{8}	1	1/8 = 0.125
(ii) an odd number	{1, 3, 5, 7}	4	4/8 = 1/2 = 0.5
(iii) a number greater than 2	{3, 4, 5, 6, 7, 8}	6	6/8 = 3/4 = 0.75
(iv) a number less than 9	{1, 2, 3, 4, 5, 6, 7, 8}	8	8/8 = 1
(v) a multiple of 3	{3, 6}	2	2/8 = 1/4 = 0.25

(i) P(8) = 1/8
(ii) P(odd) = 4/8 = 1/2
(iii) P(> 2) = 6/8 = 3/4
(iv) P(< 9) = 8/8 = 1
(v) P(multiple of 3) = 2/8 = 1/4

Why it happens: the sectors are equal in size, which is the physical reason the eight outcomes are equally likely — the arrow has as much room to stop in one sector as in any other. Part (iv) is the interesting one: every number on the wheel is less than 9, so the event is the whole sample space and P = 1, a certain event. Part (iii) is a reminder to read the inequality strictly: 'greater than 2' excludes 2 itself, giving 6 sectors, not 7.

Tip: for (v), 9 is a multiple of 3 but is not on this wheel, and 1 and 2 are not multiples of 3 — only 3 and 6 qualify. Always intersect the description with the sample space before counting.

Q10.

*A basket contains 4 red balls and 5 blue balls. One ball is drawn and laid aside, and a second ball is drawn. Draw a tree diagram to represent the possible outcomes and probabilities. Use the tree diagram to answer the following questions. (i) What is the probability of drawing a red ball and then a blue ball? (ii) What is the probability of drawing 2 blue balls?

Answer

The basket starts with 4 + 5 = 9 balls. The first ball is laid aside, so only 8 balls remain for the second draw — and which colour is missing depends on the first result. That is what the second stage of the tree has to record.

Drawing without replacement: the second-stage fractions have denominator 8, because one ball has already been laid aside.

(i) Red then blue. Multiply along that path:

P(red first) = 4/9
P(blue second | red gone) = 5/8 (5 blue still there, 8 balls left)
P(red then blue) = 4/9 × 5/8 = 20/72
= 5/18 ≈ 0.278

(ii) Two blue balls.

P(blue first) = 5/9
P(blue second) = 4/8 (only 4 blue left out of 8)
P(two blue) = 5/9 × 4/8 = 20/72
= 5/18 ≈ 0.278

Path	Working	Probability
Red, Red	4/9 × 3/8 = 12/72	1/6
Red, Blue	4/9 × 5/8 = 20/72	5/18
Blue, Red	5/9 × 4/8 = 20/72	5/18
Blue, Blue	5/9 × 4/8 = 20/72	5/18
Total	(12 + 20 + 20 + 20)/72 = 72/72	1

Why it happens: the two draws are not independent. Removing a ball changes both the total and the count of that colour, so the second-stage fractions all have denominator 8 and their numerators depend on which branch you came along. Compare Exercise Set 7.4 Q2, where the pen was put back and both stages had denominator 9 — replacement is exactly what decides whether a two-stage tree repeats itself or not.

Check it yourself: the four leaf probabilities add to 1, which is the standard test that a tree is complete. Notice that 'red then blue' and 'blue then red' come out equal at 5/18 even though the branch fractions differ: 4/9 × 5/8 and 5/9 × 4/8 are the same product written in a different order.

Q11.

*I throw a pair of 6-sided dice. Write down an event that has a probability of 0 and an outcome that has a probability of 1.

Answer

Throwing two dice gives n(S) = 6 × 6 = 36 equally likely outcomes, and the possible sums run from 2 to 12.

An event with probability 0 — anything that cannot occur:

E = 'the sum of the two dice is 1'
No pair (a, b) with a, b ≥ 1 adds to 1, so n(E) = 0
P(E) = 0/36 = 0

Other correct answers: 'the sum is 13', 'a die shows 7', 'both dice show the same number and the sum is odd'.

An event with probability 1 — anything that must occur:

F = 'the sum lies between 2 and 12 (both included)'
All 36 outcomes satisfy it, so n(F) = 36
P(F) = 36/36 = 1

Other correct answers: 'each die shows a whole number from 1 to 6', 'the sum is at least 2'.

Why it happens: P = 0 and P = 1 are the two ends of the scale, and they describe events, not luck. An impossible event has an empty favourable set; a certain event has the whole sample space as its favourable set. Everything real about two dice lies strictly between: the commonest sum, 7, has probability only 6/36 = 1/6.

Tip: the question says "an outcome that has a probability of 1", and strictly no single outcome of this experiment can have probability 1 — each of the 36 pairs has probability 1/36. What can have probability 1 is the certain event, as written above. Read it as asking for a result that is bound to happen.

Q12.

*Write the sample space and calculate the probability based on the given information. (i) Two dice are rolled. What is the probability that the sum is a prime number greater than 5? (ii) A bag contains 4 red, 3 green, and 2 blue balls. Two balls are drawn without replacement. What is the probability that both are of different colours? (iii) Three coins are tossed. What is the probability that the first coin shows heads and exactly two heads occur in total? (iv) A four-digit number is formed using the digits 1, 2, 3, and 4 with no repetition. What is the probability that the number is even? (v) A student takes a multiple-choice test with 3 questions, each having 4 options (A, B, C, D), with only one correct answer. What is the probability that the student guesses and gets exactly 2 answers correct?

Answer

(i) Two dice, sum a prime greater than 5. The sample space is all ordered pairs, n(S) = 6 × 6 = 36. Possible sums are 2 to 12; the primes among them are 2, 3, 5, 7, 11, and those greater than 5 are 7 and 11.

All 36 equally likely sums. The 6 rose cells are the sum 7 and the 2 green cells the sum 11 — together 8 of 36.

Sum 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 outcomes
Sum 11: (5,6), (6,5) → 2 outcomes
n(E) = 6 + 2 = 8
P = 8/36 = 2/9 ≈ 0.222

(ii) Two balls from 4 red, 3 green, 2 blue — different colours. There are 9 balls; drawing 2 without replacement gives

n(S) = ⁹C₂ = (9 × 8)/2 = 36 equally likely pairs
Same colour: red ⁴C₂ = 6, green ³C₂ = 3, blue ²C₂ = 1
n(same) = 6 + 3 + 1 = 10, so P(same) = 10/36 = 5/18
P(different) = 1 − 5/18 = 13/18 ≈ 0.722

Without the C notation: count the pairs directly as 4×3 (red-green) + 4×2 (red-blue) + 3×2 (green-blue) = 12 + 8 + 6 = 26 different-colour pairs, and 26/36 = 13/18. ✓

(iii) Three coins — first is heads AND exactly two heads in all.

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}, n(S) = 8
First coin H and exactly two heads: HHT, HTH → n(E) = 2
P = 2/8 = 1/4 = 0.25

Note that THH has exactly two heads but fails the first condition, so it is not counted.

(iv) Four-digit numbers from 1, 2, 3, 4 with no repetition — even.

n(S) = 4! = 4 × 3 × 2 × 1 = 24 numbers
Even ⇒ the units digit must be 2 or 4 → 2 choices
The other three digits can be arranged in 3! = 6 ways
n(E) = 2 × 6 = 12
P(even) = 12/24 = 1/2 = 0.5

(v) Three multiple-choice questions, guessing, exactly 2 correct. Each question has 4 options, so

n(S) = 4 × 4 × 4 = 64 equally likely answer-sheets
Choose which 2 of the 3 questions are right: 3 ways (QQ×, Q×Q, ×QQ)
Each right question: 1 way; the wrong one: 3 wrong options
n(E) = 3 × 1 × 1 × 3 = 9
P(exactly 2 correct) = 9/64 ≈ 0.141 or about 14%

Why it happens: every part is the same discipline — build a list of equally likely outcomes first, then count. In (i) the pairs must be ordered, because (1,6) and (6,1) are two different throws; treating the 11 sums 2…12 as the sample space would wrongly give 2/11. In (ii) the pairs are unordered, since two balls come out together; the complement is used because 'same colour' has three cases and 'different colour' has three too, but the same-colour count is smaller. In (iv) the condition sits on the units digit, so it is fixed first and the rest arranged freely. In (v) the position of the wrong answer matters, which is why the factor 3 appears.

Tip: in (iv) the answer 1/2 has a neat reason. Of the four digits, two are even and two are odd, and each digit is equally likely to occupy the units place among the 24 arrangements — so exactly half the numbers end in an even digit.

Q13.

*A box contains 4 balls numbered 1 to 4. Record a sample space using a tree diagram for the following experiments: (i) A ball is drawn, and the number is recorded. Then the ball is returned, and a second ball is drawn and recorded. (ii) A ball is drawn and recorded. Without replacing the first ball, the experimenter draws and records a second ball. (iii) What are the sizes of these two sample spaces?

Answer

(i) With replacement. The ball goes back, so all four numbers are available again at the second draw: 4 branches, then 4 more from each.

With replacement: every first number can be followed by any of the four numbers, so n(S) = 4 × 4 = 16.

S₁ = {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4),
(3,1), (3,2), (3,3), (3,4), (4,1), (4,2), (4,3), (4,4)}

(ii) Without replacement. The first ball is kept out, so the number drawn first cannot appear again: each of the 4 branches splits into only 3.

Without replacement: the first number cannot repeat, so each branch splits into only 3, giving n(S) = 4 × 3 = 12.

S₂ = {(1,2), (1,3), (1,4), (2,1), (2,3), (2,4),
(3,1), (3,2), (3,4), (4,1), (4,2), (4,3)}

(iii) Sizes.

n(S₁) = 4 × 4 = 16
n(S₂) = 4 × 3 = 12
Difference = 4, exactly the four repeats (1,1), (2,2), (3,3), (4,4)

Why it happens: replacement decides how wide the second stage of the tree is. Put the ball back and the box is unchanged, so the second draw has the same 4 choices as the first — and the pairs (k, k) are possible. Keep the ball out and one choice has been used up, so the second draw has 3 choices and no pair can repeat a number. The four missing outcomes are precisely the four repeats, which is why 16 − 12 = 4.

Tip: the order matters in both lists: (1,2) means 1 first then 2, while (2,1) is the other way round, and both are separate leaves of the tree. If the question had asked only which two numbers came out, without order, S₂ would shrink to 12 ÷ 2 = 6 unordered pairs.

Q14.

*List the elements of a sample space for the simultaneous tossing of a coin and drawing of a card from a set of 6 cards numbered 1 through 6.

Answer

The coin gives 2 results and the card gives 6, and the two happen together, so pair every coin result with every card.

S = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6),
(T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}
n(S) = 2 × 6 = 12

Coin \ Card	1	2	3	4	5	6
H	(H, 1)	(H, 2)	(H, 3)	(H, 4)	(H, 5)	(H, 6)
T	(T, 1)	(T, 2)	(T, 3)	(T, 4)	(T, 5)	(T, 6)

Why it happens: the coin result does not affect the card, and the card does not affect the coin, so no combination is ruled out and none is favoured: all 12 outcomes are equally likely at 1/12 each. This is the same list as Exercise Set 7.3 Q2(i), where a die replaced the six cards — a good sign that the structure, not the props, is what matters.

Tip: with this sample space you can answer combined questions at once. P(heads and an even card) = |{(H,2), (H,4), (H,6)}| / 12 = 3/12 = 1/4, which is also 1/2 × 1/2 — multiplying along a two-stage path.

Q15.

*Three coins are tossed, and the number of heads is recorded. Which of the following lists is a sample space for this experiment? Why do the other lists fail to qualify as a sample space? (i) {1, 2, 3} (ii) {0, 1, 2} (iii) {0, 1, 2, 3, 4} (iv) {0, 1, 2, 3}

Answer

(iv) {0, 1, 2, 3} is the sample space. With three coins the number of heads can be 0, 1, 2 or 3 — and nothing else.

List	Valid?	What is wrong
(i) {1, 2, 3}	No	Leaves out 0. TTT is a genuine result of the experiment, and it would have no outcome to belong to.
(ii) {0, 1, 2}	No	Leaves out 3. HHH can happen, so the list is incomplete.
(iii) {0, 1, 2, 3, 4}	No	Includes 4, which is impossible — only three coins are tossed, so at most three heads can appear.
(iv) {0, 1, 2, 3}	Yes	Complete and with no repeats: every toss of three coins gives exactly one of these four counts.

Why it happens: a sample space has to satisfy both rules of Section 7.3.1 at once — nothing left out and nothing that cannot occur. Lists (i) and (ii) break the first rule and list (iii) breaks the second. Adding an impossible outcome is not harmless: it would receive probability 0 and would tempt you to divide by 5.

Tip: {0, 1, 2, 3} is a correct sample space but its four outcomes are not equally likely, so P(2 heads) is not 1/4. Going back to the eight equally likely outcomes {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} gives P(0) = 1/8, P(1) = 3/8, P(2) = 3/8, P(3) = 1/8 — which add to 1. Being a valid sample space and being usable with the counting formula are two different things.

Q16.

*Suppose you drop a dye at random on the rectangular region shown in Fig. 7.8. What is the probability that it will land inside the circle with a diameter of 1 m?

Answer

From Fig. 7.8 the rectangle is 3 m by 2 m and the circle has diameter 1 m, so its radius is 0.5 m. Here the outcomes are points, not a list you can count — so probability is measured by area instead.

The dye can land anywhere in the 6 m² rectangle; it lands in the circle only over the shaded disc of area π/4 m².

Area of the rectangle = 3 × 2 = 6 m²
Radius of the circle = 1/2 = 0.5 m
Area of the circle = πr² = π × (1/2)² = π/4 m²

P(lands inside the circle) = area of circle / area of rectangle
= (π/4) ÷ 6
= π/24
≈ 3.1416/24 ≈ 0.131 or about 13.1%

Using π ≈ 22/7 gives (22/7)/24 = 22/168 = 11/84 ≈ 0.131 — the same to three decimal places.

Why it happens: "at random" here means every point of the rectangle is as likely to be hit as every other, and no region is favoured. Under that assumption the chance of landing in a part of the region is proportional to how much area that part occupies — so the counting formula n(A)/n(S) becomes area(A)/area(S). The circle covers less than an eighth of the rectangle, and the answer says so.

Tip: the answer stays π/24 wherever the circle is placed, as long as it lies wholly inside the rectangle — only the areas matter, not the position. And notice the circle covers about 13% of the sheet even though its diameter is a third of the rectangle's length: area shrinks with the square of the length.

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